WORLD FOR CIVIL

STATICS / STRENGTH OF MATERIALS - Example

In the structure shown below members ABC and BDE are assumed to be solid rigid members. Member BDE is supported by a roller at point E, and is pinned to member ABC at point B. Member ABC is pinned to the wall at point A. Member ABC is a aluminum rod with a diameter of 1 inch. For this structure:
A. Draw a Free Body Diagram showing all support forces and loads.
B. Determine the axial stress in member ABC both in section AB and section BC.
C. Determine the movement of point C due to the two applied loads.
E_al = 10 x 10⁶ psi
Unless otherwise indicated, all joints and support points are assumed to be pinned or hinged joints.

Solution:
PART A:
External support reaction - Statics:
STEP 1: Draw a free body diagram showing and labeling all load forces and support (reaction) forces, as well as any needed angles and dimensions.
STEP 2: Break any forces not already in x and y direction into their x and y components.
STEP 3: Apply the equilibrium conditions.
Sum F_x = A_x = 0
Sum F_y = A_y + E_y -16,000 lbs - 12,000 lbs = 0
Sum T_E = (16,000 lbs)(12 ft) + (12,000 lbs)(8 ft) - A_y(12 ft) = 0
Solving for the unknowns:
A_y = 24,000 lbs; E_y = 4,000 lbs

PART B:
STEP 1: Take out member ABC. Analyze force acting on it. Draw a free body diagram of ABC
STEP 2: Resolve all forces into x and y components
STEP 3: Apply equilibrium conditions.
Sum F_x = B_x = 0
Sum F_y = 24,000 lbs - B_y - 16,000 lbs = 0
Solving for the unknowns:
B_y = 8,000 lbs
Now to find the force in section AB of member ABC.

Cut member between points A and B. Look at top section (see diagram).
The internal force in section AB must be 24,000 lbs (equal to A) for equilibrium.
Stress_AB = F/A = 24,000 lbs/.785 in² = 30,600 psi

Now to find the force in section BC of member ABC.
Cut member between points B and C. Look at top section (see diagram).
The internal force in section BC must be 16,000 lbs for equilibrium.
Stress_BC = F/A = 16,000 lbs/.785 in² = 20,400 psi

PART C:
C. Def = Deformation
To find the total movement of C = Def_AB + Def_BC
move.C=[ (FL / EA)_AB + (FL / EA)_BC ]
move.C=[(24,000 lbs)(72 in)/(10*10⁶ psi)(3.14*(.5 in)²)]_AB+[(16,000 lbs)(48 in)/(10*10⁶ psi)(3.14*(.5 in)²)]_BC
move.C=(.22 in) + (.0978 in) = .318 in